Let à = − 3ỉ − 2j + 2k. Find ||ā||.

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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1.1.2

**Problem Statement: Vector Magnitude**

Let \(\vec{a} = -3\vec{i} - 2\vec{j} + 2\vec{k}\). Find \(\|\vec{a}\|\).

---

**Solution:**

The magnitude of a vector \(\vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}\) is given by the formula:

\[
\|\vec{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}
\]

For \(\vec{a} = -3\vec{i} - 2\vec{j} + 2\vec{k}\):

- \(a_1 = -3\)
- \(a_2 = -2\)
- \(a_3 = 2\)

Substitute these values into the formula:

\[
\|\vec{a}\| = \sqrt{(-3)^2 + (-2)^2 + (2)^2}
\]

\[
\|\vec{a}\| = \sqrt{9 + 4 + 4}
\]

\[
\|\vec{a}\| = \sqrt{17}
\]

Therefore, the magnitude of \(\vec{a}\) is \(\sqrt{17}\).

---

This exercise helps in understanding how to calculate the magnitude of a vector in three-dimensional space using its components.
Transcribed Image Text:**Problem Statement: Vector Magnitude** Let \(\vec{a} = -3\vec{i} - 2\vec{j} + 2\vec{k}\). Find \(\|\vec{a}\|\). --- **Solution:** The magnitude of a vector \(\vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}\) is given by the formula: \[ \|\vec{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \] For \(\vec{a} = -3\vec{i} - 2\vec{j} + 2\vec{k}\): - \(a_1 = -3\) - \(a_2 = -2\) - \(a_3 = 2\) Substitute these values into the formula: \[ \|\vec{a}\| = \sqrt{(-3)^2 + (-2)^2 + (2)^2} \] \[ \|\vec{a}\| = \sqrt{9 + 4 + 4} \] \[ \|\vec{a}\| = \sqrt{17} \] Therefore, the magnitude of \(\vec{a}\) is \(\sqrt{17}\). --- This exercise helps in understanding how to calculate the magnitude of a vector in three-dimensional space using its components.
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Find the square root of the sum of the squares of the coefficients of each unit vector.

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